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Arlecino [84]
2 years ago
9

HELP ME THIS IS FROM KA

Mathematics
2 answers:
nalin [4]2 years ago
7 0
The answer to this math problem is B
zalisa [80]2 years ago
3 0

Answer:

Line B

Step-by-step explanation:

Proportionally, it follows the graph perfectly and takes the most linear and constant path in the graph compared to the other lines.

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What is the answer to this ?
harina [27]

Answer:

The answer is option 1.

Step-by-step explanation:

In order to solve x, you have to make x the subject, by multiplying each sides by 5, to get rid of 5 on the left side.

\frac{x}{5}  = 8

\frac{x}{5}  \times 5 = 8 \times 5

x = 40

5 0
3 years ago
Ummm can someone plsss help, i need this to bring my grade up!!?
artcher [175]

Answer:

I would say it would also be 64 degrees because if you flip the HGJ to the IJG it would be the same! Hope this helps

:)

8 0
3 years ago
Read 2 more answers
the profit of a company is modeled by p = 4t + 15,500, where p is the profit and t is the time in months. what is the dependent
posledela
The dependent variable is the profit because that is what you're measuring
6 0
3 years ago
Read 2 more answers
PLASE ANSWER THIS QUESTION ​I GIVE YOU 18 POINTS
sukhopar [10]

Answer:

(a) 144

(b) 117

(c) 360

(d) 588

(e) 5472

Step-by-step explanation:

To find the LCM of two numbers, first find the prime factorization of each number. Then the LCM is the product of common and not common factors with the larger exponent.

2.

(a)

72 = 2^3 \times 3^2

144 = 2^4 \times 3^2

LCM = 2^4 \times 3^2 = 8 \times 9 = 144

(b)

39 = 3 \times 13

117 = 3^2 \times 13

LCM = 3^2 \times 13 = 117

(c)

72 = 2^3 \times 3^2

90 = 2 \times 3^2 \times 5

LCM = 2^3 \times 3^2 \times 5 = 360

(d)

84 = 2^2 \times 3 \times 7

147 = 3 \times 7^2

LCM = 2^2 \times 3 \times 7^2 = 588

(e)

152 = 2^3 \times 19

288 = 2^5 \times 3^2

LCM = 2^5 \times 3^2 \times 19 = 5472

6 0
3 years ago
PLEASE HELP! Which table shows a proportional relationship between x and y?
Inga [223]
I’m pretty sure that it is the 3 or C option.
5 0
3 years ago
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